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    a ball is hit with a velocity of 49ms^-1 at an angle x above the horizontal where sinx=3/5

    find the distance the ball travels before it hits the ground for the 1st time


    SolvedQuick Q: A particle is projected from the origin at a velocity of (5i+16j)ms^-1 where i and j are unit vectors in the horizontal and upward vertical directions respectively.

    a) find the postion vector of the particle after 2 secs

    So I said looking at the Vertical Motion

    i want to find S(V), when t=2, a=-g, u=5i+16j

    use s=ut+0.5at^2

    S(V)= 2(5i+16j)-19.6j = 10i+12.4j

    Now do I need to look at the Horizontal Motion or not

    if would just be Distance=speedxtime

    D=(5i+16j)2= 10i+32j

    What do I do now? the answer is 10i-12.4j

    b) find its distance from O after 3 secs

    isn't this just Horizontal Motion D= st

    D=(5i+16j)3=15i+48j but the answer is 15.5m ???
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    (Original post by emmaxoxo)
    i want to find S(V), when t=2, a=-g, u=5i+16j

    use s=ut+0.5at^2

    S(V)= 2(5i+16j)-19.6j = 10i+12.4j
    This is not S(V), this is the whole displacement (just S), since you have considered the whole of the initial velocity, and the whole of the acceleration. So no, you don't have to do anything else. I know it doesn't match the given answer, but I got the same, so either we're both making the same mistake, you made a typo, or the book is wrong.

    For b, you need to do the same procedure as in a, but using t=3. Then when you have s=ai+bj, you can work out the distance.
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    (Original post by meatball893)
    This is not S(V), this is the whole displacement (just S), since you have considered the whole of the initial velocity, and the whole of the acceleration. So no, you don't have to do anything else. I know it doesn't match the given answer, but I got the same, so either we're both making the same mistake, you made a typo, or the book is wrong.
    Ok, so how did I get +ve 12.4j when it's suppose to be -Ve ?

    Soory, just got your edited.

    What about B)
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    Maybe the book has a mistake. I can't see how it got -12.4j.
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    (Original post by meatball893)
    Maybe the book has a mistake. I can't see how it got -12.4j.
    Thank you, I've got it part b now aswell. Yh I'll just leave it as wrong answer in the book

    Can you help me on this new question?

    a ball is hit with a velocity of 49ms^-1 at an angle x above the horizontal where sinx=3/5

    find the distance the ball travels before it hits the ground for the 1st time

    do you find the time first, then using that time, find the distance? if soo I'm getting it wrong

    t=3 d=117.6 whereas d=235m
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    Well again use s=ut+\frac12 at^2 , but only consider the vertical component of each vector. So s=0, because the ball has returned to the same level. Are you happy that u_v=49\sin x = \frac{147}5 ?
    Anyway, solve the equation to find t, then consider the horizontal components. Since there is no horizontal acceleration, the equation is just s_h=u_ht=49t\cos x.
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    (Original post by meatball893)
    Well again use s=ut+\frac12 at^2 , but only consider the vertical component of each vector. So s=0, because the ball has returned to the same level. Are you happy that u_v=49\sin x = \frac{147}5 ?
    Anyway, solve the equation to find t, then consider the horizontal components. Since there is no horizontal acceleration, the equation is just s_h=u_ht=49t\cos x.
    Thank you, I've got it. In this case do we not know what v= ?
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    (Original post by emmaxoxo)
    Thank you, I've got it. In this case do we not know what v= ?
    What do you mean by v?
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    (Original post by meatball893)
    What do you mean by v?
    I trying to understand more, as I said for that Q that t=? u=49sinx a=-g and v=0, whereas it should have been s=0

    so i ended up using v=u+at, how do you know to use s=0 (i get it's 0 because it when up and back down again), and not v=0?

    Also another Q;

    a ball is thrown from o with a speed of 30ms^-1, 30 degrees from the horizontal, given that it clears the top of a wall, 20m from O, find the height of the wall?

    I first found t, which i got as 2/3. the used s=ut+0.5at^2 getting s=7.82m whereas it should be 8.64m ?
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    (Original post by emmaxoxo)
    I trying to understand more, as I said for that Q that t=? u=49sinx a=-g and v=0, whereas it should have been s=0

    so i ended up using v=u+at, how do you know to use s=0 (i get it's 0 because it when up and back down again), and not v=0?
    When you throw a ball in the air, and it falls back to ground, it its velocity zero just before it hits the ground? No, but its vertical displacement is, because it has returned to the same level (well obviously your arm isn't at the same level as the ground, but you get the idea).
    Since s=0 (as it has returned to the same level), you can use s=ut+\frac12 at^2 to find t.

    Also another Q;

    a ball is thrown from O with a speed of 30ms^-1, 30 degrees from the horizontal, given that it clears the top of a wall, 20m from O, find the height of the wall?

    I first found t, which i got as 2/3. the used s=ut+0.5at^2 getting s=7.82m whereas it should be 8.64m ?
    First find the time at which the ball passes over the wall. You consider the horizontal motion, so there is no acceleration and you can use s=ut \iff t=\frac su . This gives you t.
    [I think what you've done wrong here is you have used the whole velocity, whereas you want the horizontal component: u_h=30\cos\theta].
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    (Original post by meatball893)
    When you throw a ball in the air, and it falls back to ground, it its velocity zero just before it hits the ground? No, but its vertical displacement is, because it has returned to the same level (well obviously your arm isn't at the same level as the ground, but you get the idea).
    Since s=0 (as it has returned to the same level), you can use s=ut+\frac12 at^2 to find t.



    First find the time at which the ball passes over the wall. You consider the horizontal motion, so there is no acceleration and you can use s=ut \iff t=\frac su . This gives you t.
    [I think what you've done wrong here is you have used the whole velocity, whereas you want the horizontal component: u_h=30\cos\theta].
    Sorry, another question;

    I think I have just got the +'s and -'s mixed up here.

    a plane is climbing at an angle of 2, while ,maintaining a speed of 400ms^-1, a package is released and travels a horizontal distance of 2500m before hitting the ground. the package is the particle, with the same initail velcity as the plane.

    find the height of the plane above the ground, the moment the package was released.

    I worked out t=6.25, (d=st)

    u=400sin2 t=6.25 s=? a+-g

    s=ut+0.5at^2

    (Rounded) s=87-191=-104, it's suppose to be 104

    therefore it needed to be -87+191=104, which signs have i got the wrong way round and why?
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    No, your answer is correct. Remember, the formula is for displacement, not distance. Since you've taken upwards as positive, any motion downwards is negative. So the displacement is -104, but the distance is |-104|=104.
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    (Original post by meatball893)
    No, your answer is correct. Remember, the formula is for displacement, not distance. Since you've taken upwards as positive, any motion downwards is negative. So the displacement is -104, but the distance is |-104|=104.
    Oh yeah, can you help me on this one again please, I still don't get it

    Also another Q;

    a ball is thrown from O with a speed of 30ms^-1, 30 degrees from the horizontal, given that it clears the top of a wall, 20m from O, find the height of the wall?

    I first found t, which i got as 2/3. the used s=ut+0.5at^2 getting s=7.82m whereas it should be 8.64m ?

    First find the time at which the ball passes over the wall. You consider the horizontal motion, so there is no acceleration and you can use s=ut \iff t=\frac su . This gives you t.
    [I think what you've done wrong here is you have used the whole velocity, whereas you want the horizontal component: u_h=30\cos\theta].
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    (Original post by emmaxoxo)
    Oh yeah, can you help me on this one again please, I still don't get it
    You went wrong when you did t=\frac sv =\frac {20}{30} .
    It should have been t=\frac{20}{30\cos 30^\circ} , because you are considering horizontal motion here.
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    (Original post by meatball893)
    You went wrong when you did t=\frac sv =\frac {20}{30} .
    It should have been t=\frac{20}{30\cos 30^\circ} , because you are considering horizontal motion here.
    Thank you soo mcuh for your help. Got it now!
 
 
 
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